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Proof step 5 from proof inj (previous step)

n→i(m) = n→i(n0) + 1, n→i(m -1) = n→i(n0) → m -1 = n0, ∀ m. n→i(m) = n→i(n0) → m = n0 ⊦ m = n0 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n→i(m) = n→i(n0) + 1, ∀ m. n→i(m) = n→i(n0) → m = n0, n→i(m - 1) = n→i(n0) → m - 1 = n0 ⊦ m = n0 + 1

Rule argument

Additional information

Used simplifier rules: pred-03: ⊦ n -1 = n - 1; from specification nat